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I am having difficulty finding the radius of convergence of

$e^{-x^{2}}$

this is for introductory analysis course. Have looked at even and odd subsequences of powerseries, but so far unable to put the pieces together. Any help appreciated.

Started with:
$e^{-x^{2}} = \Sigma_{k=0}^\infty \frac{(-x)^{2k}}{k!}= \Sigma_{n=0}^\infty S_nx^n$
s.t. $S_n = 0$
when n odd and
$S_n = \frac{(-1)^{n}}{\frac{n}{2}!}$ when n even.

Then $S_{2n+1} = (0,0,0,0,...)$, converging to $0$, and $S_{2n} = \frac{1}{n!}$, and with ratio test we get

$lim_{n \rightarrow \infty}\left|\frac{\frac{1}{(n+1)!}}{\frac{1}{n!}}\right| = lim_{n \rightarrow \infty}\frac{1}{n+1} = 0$

Jebus
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  • are you sure the $e^{-x^{2}} = -\Sigma_{k=0}^\infty \frac{x^{2k}}{k!}= -\Sigma_{n=0}^\infty S_nx^n$ because $e^{-x^{2}}$ is always positive while $-\Sigma_{k=0}^\infty \frac{x^{2k}}{k!}= -\Sigma_{n=0}^\infty S_nx^n$ is always negative – Kamster Mar 18 '15 at 19:53
  • corrected, I think – Jebus Mar 18 '15 at 19:59
  • still negative notice the difference between $-x^{2k}$ and $(-x)^{2k}$ – Kamster Mar 18 '15 at 20:00
  • thanks. any input on logic/methodology errors, or simply the notational? – Jebus Mar 18 '15 at 20:02
  • for ratio should have an $x$ somewhere in there and if it ends up converging to zero for any $x$ then radius of convergence is infinity i.e. interval of convergence is $\mathbb{R}$ – Kamster Mar 18 '15 at 20:04
  • like this? $lim_{n \rightarrow \infty}\left|\frac{\frac{1}{(n+1)!}x^{n+1}}{\frac{1}{n!}x^n}\right| = |x|lim_{n \rightarrow \infty}\frac{1}{n+1} = |x|0$ and if so, is it enough to show both even and odd subsequences converging to 0? – Jebus Mar 18 '15 at 20:11
  • if original sequence converges to zero you have any subsequence converges to zero. Im a little confused why you want to prove that to prove radius of convergence though – Kamster Mar 18 '15 at 21:02

1 Answers1

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Since we know that $$e^x=\sum_{n=0}^{\infty}\frac{(x)^n}{n!}$$ we then have $$e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}$$ Thus by ratio test we see $$\lim_{n\rightarrow\infty}\left|\frac{\frac{(-1)^{n+1}x^{2(n+1)}}{(n+1)!}}{\frac{(-1)^nx^{2n}}{n!}}\right|=\lim_{n\rightarrow\infty}\left|\frac{x^2}{n+1}\right|=0$$ Thus the radius of convergence is $\infty$ and interval of convergence is $\mathbb{R}$

Kamster
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